Math Problem Statement

Min Z = 25x+15y+410 subject to X+y<=100 , x<=60 , y<=50 and X+Y>=60

Solution

Given your linear programming problem, the objective is to minimize the function Z=25x+15y+410Z = 25x + 15y + 410 subject to a set of constraints. Here's the full formulation:

Objective Function:

Minimize Z=25x+15y+410\text{Minimize } Z = 25x + 15y + 410

Subject to the Constraints:

  1. x+y100x + y \leq 100
  2. x60x \leq 60
  3. y50y \leq 50
  4. x+y60x + y \geq 60

Steps to Solve:

  1. Identify the feasible region: Plot the constraints on a graph with xx on the horizontal axis and yy on the vertical axis. The feasible region is the area where all constraints overlap.

  2. Find the corner points of the feasible region: These are the points where the constraint lines intersect. The optimal solution in linear programming occurs at one of these corner points.

  3. Calculate the value of ZZ at each corner point: Substitute the coordinates of each corner point into the objective function to find which one gives the minimum value.

  4. Determine the minimum value: The corner point with the lowest value of ZZ is the optimal solution.

Let's solve it:

1. Constraints:

  • x+y100x + y \leq 100
  • x60x \leq 60
  • y50y \leq 50
  • x+y60x + y \geq 60

2. Find intersection points:

  • Intersection of x+y=100x + y = 100 and x=60x = 60 gives (60,40)(60, 40).
  • Intersection of x+y=100x + y = 100 and y=50y = 50 gives (50,50)(50, 50).
  • Intersection of x+y=60x + y = 60 and x=60x = 60 gives (60,0)(60, 0).
  • Intersection of x+y=60x + y = 60 and y=50y = 50 gives (10,50)(10, 50).

3. Evaluate ZZ at each point:

  1. At (60,40)(60, 40): Z=25(60)+15(40)+410=1500+600+410=2510Z = 25(60) + 15(40) + 410 = 1500 + 600 + 410 = 2510
  2. At (50,50)(50, 50): Z=25(50)+15(50)+410=1250+750+410=2410Z = 25(50) + 15(50) + 410 = 1250 + 750 + 410 = 2410
  3. At (60,0)(60, 0): Z=25(60)+15(0)+410=1500+0+410=1910Z = 25(60) + 15(0) + 410 = 1500 + 0 + 410 = 1910
  4. At (10,50)(10, 50): Z=25(10)+15(50)+410=250+750+410=1410Z = 25(10) + 15(50) + 410 = 250 + 750 + 410 = 1410

4. Minimum Value:

The minimum value of ZZ occurs at (10,50)(10, 50) where Z=1410Z = 1410.

Optimal Solution:

  • Optimal point: x=10x = 10, y=50y = 50
  • Minimum value of ZZ: 1410

Would you like more details on this solution, or do you have any questions?

Related Questions:

  1. How can I graphically represent this linear programming problem?
  2. What is the significance of the constant 410 in the objective function?
  3. How does the feasible region affect the optimal solution?
  4. Can we solve this problem using the Simplex method instead of graphically?
  5. What if one of the constraints were non-linear? How would that change the approach?

Tip:

When solving linear programming problems graphically, always check the boundaries of your feasible region carefully, as they determine where the optimal solution lies.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Constraints

Formulas

Objective Function for Linear Programming

Theorems

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Suitable Grade Level

Advanced High School